06 finite difference-6
DESCRIPTION
Heat transfer (finite difference method)TRANSCRIPT
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Numerical MethodFinite Differencing
x = 1
1 2 3 i-1 i i+1 N-2 N-1 N
x = 0 h
Consider uniform interval h is constant
f(x) is assigned at each node, f1, f2, fi-1, fi, fi+1, fN-2, fN-1, fN
nodeeachatxf
xf
eApproximat 2
2
&
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)(Accuracy2
)(21
)(21
)12(NodesInternalFor:ExpansionSeriesTaylor
sideeachatnodeoneuse:Example
311
322
2
1
322
2
1
hOhff
xf
xf
forSolve
hOhxf
hxf
ff
hOhxf
hxf
ff
Ni
ii
i
i
iiii
iiii
Discretization
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)(Accuracy2
)(21
)(21
)12(NodesInternalFor:ExpansionSeriesTaylor
sideeachatnodeoneuse:Example
42
112
2
2
2
322
2
1
322
2
1
hOh
fffxf
xf
forSolve
hOhxf
hxf
ff
hOhxf
hxf
ff
Ni
iii
i
i
iiii
iiii
Discretization
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)(Accuracy234
Eliminate
)(4212
)(21
0orknownis,specifiedisGradient)(
knownis:boundarytheatspecifiedFunction)(
3
1132
12
2
32
12
2
113
32
12
2
112
111
1
hOhxf
fff
xf
hOhxf
hxf
ff
hOhxf
hxf
ff
Bfxf
Axf
b
fa
For Nodes at the Boundary Nodes (i=1, i=N)
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Example: Numerical Solution of the Fin Problem
0)1()1(&1)0(
)()(
1000
2
H
specifiedareXfandXawhere
XfAd
a
x = 1
1 2 3 i-1 i i+1 N-2 N-1 N
x = 0 h
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For Interior Nodes (2 ≤ i ≤ N-1)
12
02
2
2&2
eapproximatweside,eachatnodeoneUsing
0
211
211
21111
Nifor
fAd
ha
ha
hh
iiii
iiii
i
iiii
iii
x = 1
1 2 3 i-1 i i+1 N-2 N-1 N
x = 0 h
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Boundary Nodes (i = 1 and i = N)
NNN
N
NNNNNN
N
NNNN
NNNN
Hh
Hh
hOhh
hOhh
NiForiFor
234obtain,Eliminate
0since234Eliminate
)(4212
)(21
1,1
21
21
322
321
1
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Algebraic Equation
matrixt coefficien
0...,0,0,2/22/2
vector known :,1...,3,2
vector unknown ::formmatrixtheIn
:11
xNNA
Thahab
b
TNNx
xbAx
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Algebraic Equation
hHAAA
hiahiaA
AifdhiaA
hiahiaA
hahaA
AfdhaA
NNNNNN
ii
ii
ii
2341
2/2/
0/22/2
2/2/
2/22/2
0/222/22
:formmatrix Aoft coefficienThe
1,12,13,1
1,
,
1,
2,1
1,1
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Solution
solution Iterative - MethodIndirect
1matrixInvertMethodDirect
:solveTo
bAx
A
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ASSIGNMENT 2 – due Feb 16 Fin
w0Tb
T0
k=k11
k=k22
Streaming fluid
r1 r2r3
h=h1h=h2
r1
r2
r3
Top view
h=0
Cross-section
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ASSIGNMENT 2
For two fin whose cross - sections are shown calculate the temperature and the heat flux at any cross -section. Run the program for a perfect interface.
Find the temperature at the cross-section where the fin changes shape. Plot the temperature as a function of r in r1 ≤r ≤ r3.
.15,0.9,0.7,5,0.3,/60,/40
,/150,/100,20,150
3210
21
22
210
cmrcmrcmrcmwCmWkCmWk
CmWhCmWhCTCTb
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ASSIGNMENT 2
Use the fact that the temperature and the heat flux are continuous at the cross-section for a perfect interface.
xTk
xTk
tzyxTtzyxT II
22
11
21 ),,,(),,,(
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Conduction-Conditions at the Interface
Perfect Interface
xTk
xTk
tzyxTtzyxT II
22
11
21 ),,,(),,,(
Imperfect Interface
x = xI
Material 2k2 T2(x,y,z)
Material 1k1 T1(x,y,z)
Interface
xTk
xTk
xTkTThc
22
11
1121Rc is the contact resistance
Rc ≈ 10-6 – 10-3 m2K/W
hc ≈ 102 – 106 W/m2K
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RkhrRfR
AARa
RTTTTRfkSh
wrArwArrR
b
1
111
0
11
1001
1111
010011
4)( and)(
).()()(
2,2,/:1Region
Introduce Normalized Variables:
10
2112
11211
1
1222
2 where0)(
:)(for Equation where1 :1Region
kwrhRR
RrrRRR
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Introduce Normalized Variables:
0
1
2
2122
0
22
2002
2222
2021
13332
tan2with
4)( and1)(
).()()(
tan2,2,/ where:2Region
wrc
RkhrRfRRcR
AARa
RTTTTRfkSh
rrwwrwArrRrrRRRR
b
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0
122
20
2122
222222
2
32
tan2 where1)(
2 where0)(
:)(for Equation :2Region
wrcRRcRRa
kwrhRa
RRRR
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0)( tipat the
whereinterface at the continuousflux heat and
)continuous re(temperatuinterface, at the
1)1(1base at the
323
212221
2221
2
1
RRR
kkRR
RRRR
R
Boundary Conditions – with the Perfect Interface
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0)(tip theat
whereinterface theat continuousflux heat and
where0interface, theat
1)1(1base theat
323
212221
11c211
2
1
RRR
kkRR
krhHHRR
R
cc
Boundary Conditions – with the imperfect Interface
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212221
2221
321
20
2122
22222
32
10
2112
11211
2
where
interfaceperfect aFor 0)(,1)1(
2 where0)(
For
2 where0)(
1 For
kkRRRR
RkwrhRR
RRRkwrhRR
RR
Special Case = 0
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:conditionsboundary theapplyingbydeterminedbecan),,,(Constants
)(and
)(kind second andfirst zero,order of
funtions Bessel modifiedin expressed becan )( and )(for solution The
20202
10101
21
DCBARDKRCIR
RBKRAIR
RR
Solution
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2212212
2112111
220220
210210
321321
1010
and,
,0,1
:are for Equations
RDKRCIRBKRAI
RDKRCIRBKRAI
RDKRCIBKAI(A,B,C,D)
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hiRkwrh
RRcRaN-iI
hiRkwrh
hiRa-Ii
Rh
ah
a
i
iii
i
ii
iiii
iiii
i
21with2
,1 11for
11with2
11 12for
02
2nodesinterior for ng,differenci centralUsing
20
2122
22
10
2112
1
211211
R = R3
1 2 3 I1-1 (I1,I2) I2+1 N-2 N-1 N
R = 1 h R = R2
Finite Difference Discretization – for each regionUse two nodes at the interface one for region 1 (I1) and one for region 2 (I2)
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Boundary condition: Nodes i = 1 and i = N
034 insulated is Tip
1,11 base at the re temperatuSpecified
21
1
NNNNiForNi
iFori
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Condition at the interface: Interface Nodes, i = I1 ,I2
(I1 ,I2 are at the same point)
hh
h
h
IIIIII
III
III
II
243
243
243
243
)continuousheatflux (
)continuous eTemperatur(:interfaceperfect aFor
1212
12I
12I
II
222111
222
2
111
1
21
21
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Condition at the interface: Interface Nodes (i = I1 ,I2)
hh
Hh
H
IIIIII
IIcIII
IIc
243
243
)continuousflux heat (
02
43
)resistancecontact (0:interfaceimperfect an For
1212
II
12
I
222111
21
21
111
211
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22312
232321
11
1IIRegion For IRegion For
and11
dd
RRdRd
dd
RdRd
RRRRRR
Another Approach: Multi-region Problems - Transformation
R
1 R2 R3
0 01
ξ1
ξ2
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variablesdependent areandst variableindependen areand
10 For
0110 For
0111
1
21
21
2
22323222222
23
1
121211112
2
RRRaRR
RaR
Multi-region Problems – Transformation
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,0)0(,1)0(
01
0111
110 For
21
223322222
23
1221112
2
with
RRRaRR
RaR
Multi-region Problems – Final
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)1(1)1(1
1
0)1()1()1(1
1:interfaceimperfect an For
)1(1)1(1
1)1()1(
:interfaceperfect aFor
223
12
2112
223
12
21
RRR
HR
RRR
c
Multi-region Problems – Final